    Question

# An ideal monoatomic gas undergoes different types of processes which are described in column I. Match the corresponding statements in column II. The parameters involved have standard usual meaning. Column IColumn II(A)P=2V2(p) If volume increases thentemperature will also increase(B)PV2=constant(q) If volume increases thentemperature will decrease(C)C=CV+2R(r) For expansion, heat will haveto be supplied to gas(D)C=CV−2R(s) If temperature increases then theworkdone by gas is positive(t) If temperature decreases then workdone by gas is positive.

A
Ap,r,s Bq,t Cp,r,s Dq,r,t
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B
Ap,t Bq,r Cq,s Dq,r,s
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C
Aq,t Bp,r,s Cp,r,s Dq,r,s
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D
Aq,r,s Bq,t Cp,r,s Dp,r,t
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Solution

## The correct option is A A→p,r,s B→q,t C→p,r,s D→q,r,tIf P=2V2, then using ideal gas equation, PV=nRT ⇒(2V2)(V)=nRT or, nRT=2V3 or, T∝V3 Hence if volume ↑, temperature of gas will also increase. Increase in temperature means dT is +ve. or, dU∝dT ∴dU is also +ve. Using first law of thermodynamics; dQ=dU+dW ⇒dQ is also +ve (∵dW is +ve for expansion of gas (V↑)) Hence heat is supplied during expansion of gas. ∴(A)→p,r,s (B) If PV2=constant, using ideal gas equation PV=nRT ⇒(nRT)V=constant VT=constant V∝1T Hence if volume of gas increases, its temperature will decrease. For polytropic process, C=CV+R1−x C=3R2−R=R2 Heat supplied is dQ=nCdT If temperature decreases dQ is negative Now, for rise in temperature (dT→+ve) the volume decrease (V↓) or workdone by gas is -ve. However the heat supplied to gas in this case. dQ∝dT dT→+ve means dQ is also +ve. ⇒(B)→q,t (C) Given, C=CV+2R For polytropic process, C=CV+R1−x 1−x=0.5 x=0.5 P√V=Constant Using PV=nRT T√V=constant Hence with increase in temperature, V↑ Also if V↑ then dW is +ve and dU is +ve (T↑) From relation (i), dQ=dU+dW dQis +ve, if dU & dW are positive (when temperature of gas increases) Hence heat is supplied (dQ→+ve) when gas is expanding [(V↑) or dW is +ve] ⇒(C)→p,r,s (D) Given, CV−2R=C C=CV+R1−x 1−x=−0.5 x=1.5 PV1.5=Constant T√V=Constant Hence with increase in temperature, the volume of gas decreases. ⇒T↑(dT is +ve) then V↓ (compression of gas), dW is -ve. Also, dQ=n(CV−2R)dT For monoatomic gas CV=32R i.e, C<0 Thus dQ is -ve when temperature increases, or we can say heat is supplied to gas (dQ→+ve) when its temperature decrease (T↑) or volume increases. ∴(D)→q,r,t Why this question?Tip: It tests your analytical skills in relatingthe workdone by gas, change in temperature &volume by correlating the ideal gas equation(PV =nRT) with specific heat capacity of gasand first law of thermodynamics (dQ = dU + dW)  Suggest Corrections  2      Explore more