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Question

# 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is (a) 30 calories (b) 50 calories (c) 70 calories (d) 90 calories

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Solution

## (b) 50 calories It is given that 70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. Also, specific heat at constant pressure, ${C}_{\mathrm{p}}=\frac{∆Q}{n∆T}\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{p}}=\frac{70}{2×\left(35-30\right)}\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{p}}=\frac{70}{2×5}\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{p}}=7\mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$ For an ideal gas, ${\mathrm{C}}_{\mathrm{p}}-{\mathrm{C}}_{\mathrm{v}}=\mathrm{R}=8.314\mathrm{J}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\simeq 2\mathrm{calories}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{v}}={C}_{\mathrm{p}}-R\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{v}}=\left(7-2\right)\mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{v}}=5\mathrm{calories}-{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1}\phantom{\rule{0ex}{0ex}}⇒{C}_{\mathrm{v}}=\frac{∆Q}{n∆T}\phantom{\rule{0ex}{0ex}}⇒5=\frac{∆Q}{2×\left(35-30\right)}\phantom{\rule{0ex}{0ex}}⇒∆\mathrm{Q}=5×2×\left(35-30\right)\phantom{\rule{0ex}{0ex}}⇒∆\mathrm{Q}=5×2×5\phantom{\rule{0ex}{0ex}}⇒∆\mathrm{Q}=50\mathrm{calories}\phantom{\rule{0ex}{0ex}}$ Therefore, 50 calories need to be supplied to raise the temperature of 2 moles of gas from 30-35 oC at constant volume.

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